JEE Mathematics Continuity and Differentiability MCQs Set 20

Answer: (c) 2

Explanation:
1. \( f(f(x)) \) can fail to be continuous only where \( f \) is discontinuous (\( x=1,2 \)) or where \( f(x) \) equals a discontinuity point of \( f \) (\( f(x)=1 \) gives \( x=0,3 \); \( f(x)=2 \) gives \( x=2 \)).
2. Candidates: \( x=0,1,2,3 \).
3. \( x=0 \): \( f(f(0))=f(1)=0 \) and for small \( x\gt 0 \) \( f(x)\to1^- \), \( f(f(x))\to0 \): continuous.
4. \( x=1 \): \( f(f(1))=f(0)=1 \); \( f(x)\to0^+ \) from the left and \( 3^+ \) from the right, both giving \( f(f(x))\to1 \): continuous.
5. \( x=2 \): \( f(f(2))=f(2)=2 \) but for \( x\to2^- \) \( f(x)\to4^- \) and \( f(f(x))\to0 \): discontinuous.
6. \( x=3 \): \( f(f(3))=f(1)=0 \) but for \( x\to3^- \) \( f(x)\to1^+ \) and \( f(f(x))\to3 \): discontinuous.
7. So there are exactly 2 points, \( x=2,3 \).

Teacher's Note:
1. List the points where the inner function or its value hits a discontinuity of f.
2. Test each candidate with one-sided limits and the value.

Answer: (a) 3

Explanation:
1. \( t=\frac1{x-1} \) is discontinuous at \( x=1 \).
2. \( t^2+t-2=(t+2)(t-1) \) vanishes at \( t=-2 \) and \( t=1 \).
3. \( \frac1{x-1}=-2\Rightarrow x=\frac12 \) and \( \frac1{x-1}=1\Rightarrow x=2 \).
4. So y is discontinuous at \( x=1,\frac12,2 \): 3 points.

Teacher's Note:
1. Find where the inner function and the outer function each fail.
2. Convert the bad values of t back into values of x.

Answer: (d) \( x=0 \)

Explanation:
1. \( \operatorname{sgn}(x)=-1,0,1 \) for \( x\lt 0,x=0,x\gt 0 \).
2. It is constant on each side of 0 and jumps at 0.
3. So it is discontinuous only at \( x=0 \).

Teacher's Note:
1. Look at the one-sided limits at 0: -1 and 1.
2. Elsewhere the function is constant.

Answer: (b) \( f(x)=ax^2+bx \)

Explanation:
1. \( \frac{|x|}x \) tends to \( 1 \) from the right and \( -1 \) from the left.
2. So the right limit is \( f(0) \) and the left limit is \( -f(0) \).
3. The limit exists only if \( f(0)=0 \).
4. Only \( ax^2+bx \) always satisfies \( f(0)=0 \).

Teacher's Note:
1. The sign factor flips across 0, so f(0) must be 0.
2. Check which option forces f(0) = 0.

Answer: (a) \( (4,-1) \)

Explanation:
1. On \( (0,4) \): \( [x]=0,1,2,3 \) and \( \sin0=0 \), \( \sin1,\sin2,\sin3\in(0,1) \), so \( [\sin[x]]=0 \).
2. On \( [4,2\pi) \): \( [x]=4,5,6 \) and \( \sin4,\sin5,\sin6\in(-1,0) \), so \( [\sin[x]]=-1 \).
3. f jumps from 0 to -1 at \( x=4 \), so \( (4,-1) \) is the point of discontinuity (f(4) = -1).

Teacher's Note:
1. Evaluate [x] on each unit interval and then sin and the outer bracket.
2. Look for the point where the value changes.

Answer: (c) no points of discontinuity

Explanation:
1. \( x \) and \( x-x^2 \) are polynomials, hence continuous.
2. The modulus of a continuous function is continuous.
3. So f is continuous everywhere on \( [-1,2] \).

Teacher's Note:
1. Modulus preserves continuity.
2. Differences of continuous functions are continuous.

Answer: (b) it is discontinuous for \( x=n^{1/3} \), where n is any integer

Explanation:
1. \( [x^3] \) jumps exactly where \( x^3=n \), i.e. \( x=n^{1/3} \), and the parity of \( [x^3] \) changes there, so f jumps between 1 and -1.
2. Option 2: \( f(3/2)=(-1)^{[3.375]}=(-1)^3=-1 \): false.
3. Option 3: on \( (-1,0) \) f = -1 and on \( [0,1) \) f = 1 (constants), so \( f'(x)=0 \): false.
4. Option 4 is false by the jumps.

Teacher's Note:
1. Check each option separately.
2. Parity flips every time [x^3] increases by 1.

Answer: (d) \( a\lt -2 \) or \( a\gt 2 \)

Explanation:
1. sgn is discontinuous where its argument is zero or changes sign.
2. So \( 2\sin x+a \) must never vanish: \( \sin x\ne-\frac a2 \) for all x.
3. That needs \( \left|\frac a2\right|\gt 1 \), i.e. \( a\lt -2 \) or \( a\gt 2 \).

Teacher's Note:
1. The range of sin x is [-1, 1].
2. Make -a/2 lie outside it.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( [x] \) is an integer, so \( \sin([x]\pi)=0 \) for every x.
2. Hence \( f(x)=0 \) for all x: a constant function, continuous on R. Assertion true.
3. Reason is true and it is exactly why f is continuous (f is the constant function 0).

Teacher's Note:
1. Evaluate sin(n pi) = 0 for integer n.
2. The function collapses to the zero function.

Answer: (b) p-2, q-4, r-5, s-3

Explanation:
1. (p) \( \log|x| \) is undefined at 0 and zero at \( \pm1 \): 3 points. (p to 2)
2. (q) \( (1+x)^{1/\sin x}\to e \). (q to 4)
3. (r) \( \frac{\cos x-\sin x}{\cos^2x-\sin^2x}=\frac1{\cos x+\sin x}\to\frac1{\sqrt2} \). (r to 5)
4. (s) \( \frac{2\sin x}{x(\sqrt{1+\sin x}+\sqrt{1-\sin x})}\to\frac22=1 \). (s to 3)

Teacher's Note:
1. Evaluate each limit separately, then read off the pairing.
2. Rationalise or factor where needed.

Answer: (c) A is true, but R is false.

Explanation:
1. \( \left|x^2\sin\frac1x\right|\le x^2\to0 \), so f is continuous at 0 and A is true.
2. \( h(x)=x^2 \) is continuous, but \( \lim_{x\to0}\sin\frac1x \) does not exist, so g is not continuous at 0. R is false.

Teacher's Note:
1. Use the squeeze theorem for the assertion.
2. sin(1/x) oscillates between -1 and 1 near 0.

Answer: (a) 1

Explanation:
1. For \( 0\lt x\lt 1 \): \( [x]=0 \), so the right limit is \( \alpha \). Continuity: \( \alpha=2 \).
2. For small \( x\lt 0 \): \( \frac{\sin x-x}{x^3}=-\frac16+\frac{x^2}{120}-\cdots \), slightly greater than \( -\frac16 \), so its integral part is \( -1 \).
3. Left limit: \( \beta-1=2 \) gives \( \beta=3 \).
4. \( \beta-\alpha=1 \).

Teacher's Note:
1. Use the series sin x = x - x^3/6 + x^5/120.
2. The bracket value is -1 because the quantity is just above -1/6.

Answer: (b) \( (-2,-1) \)

Explanation:
1. For the left limit to be finite, \( a+2=0 \), so \( a=-2 \).
2. Then \( \frac{2\cos x-2}{x^2}\to-1 \).
3. At \( x\ge0 \) near 0: \( [x+4]=4 \), so \( f(0)=b\tan\frac\pi4=b \).
4. Continuity gives \( b=-1 \).

Teacher's Note:
1. The numerator must vanish at 0 for a finite left limit.
2. tan(pi/4) = 1.